O.k. I have a question in which I am given that a function is 'defined' on an interval [a,b], for

I tried to look for the exact definition of what it means for a function to be 'defined' on an interval but I couldn't find it.

Can it be assumed that the function is bounded if we are given that it is 'defined', and what is the exact definition of a function be 'defined' on an interval?

January 24th 2009, 09:29 PM

Jhevon

Quote:

Originally Posted by Jen

O.k. I have a question in which I am given that a function is 'defined' on an interval [a,b], for

I tried to look for the exact definition of what it means for a function to be 'defined' on an interval but I couldn't find it.

Can it be assumed that the function is bounded if we are given that it is 'defined', and what is the exact definition of a function be 'defined' on an interval?

Hey Jen, long time no see.

things like boundedness or continuity and such can't be assumed

defined on an interval simply means defined for each point in that interval. meaning you can plug in any point in that interval and the function makes sense (no division by zero etc)

January 24th 2009, 09:31 PM

Jen

But wouldn't that imply boundedness? because then the function couln't shoot off to infinty if it is defined on the interval right?

January 24th 2009, 09:32 PM

Jhevon

Quote:

Originally Posted by Jen

But wouldn't that imply boundedness? because then the function couln't shoot off to infinty if it is defined on the interval right?

right

continuity is not given though :p

January 24th 2009, 09:33 PM

Jen

So then I can assume that it is bounded if it is defined on the closed interval?

January 24th 2009, 09:34 PM

Jhevon

Quote:

Originally Posted by Jen

So then I can assume that it is bounded if it is defined on the closed interval?

yup.

is this an integration question?

January 24th 2009, 09:34 PM

Jen

Oh, well I don't need continuity to use the theorem that I need though. (Happy)

January 24th 2009, 09:36 PM

Jen

Yeah, I am supposed to prove that if a function (f) is integrable on the closed interval [a,b] and g is defined on the interval and g(x)=f(x) at all points on the interval but some c in [a,b] then they still have the same integral.

January 24th 2009, 09:41 PM

Jen

So, off the top of my head, I am thinking that if I use the same partition that works in making f integrable and take a refinement of it by adding some "small" interval around the point c then I can show that they are the same.

(in a nutshell, lol)

January 24th 2009, 09:57 PM

Jhevon

Quote:

Originally Posted by Jen

So, off the top of my head, I am thinking that if I use the same partition that works in making f integrable and take a refinement of it by adding some "small" interval around the point c then I can show that they are the same.

(in a nutshell, lol)

indeed, that is one approach. let me know how it works out.

(of course, there is a general statement for this: we can have f(x) and g(x) disagreeing at finitely many points, not just one, and it would still be true)